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%I #27 May 17 2021 17:22:27
%S 1,4,2,12,12,4,32,48,32,8,80,160,160,80,16,192,480,640,480,192,32,448,
%T 1344,2240,2240,1344,448,64,1024,3584,7168,8960,7168,3584,1024,128,
%U 2304,9216,21504,32256,32256,21504,9216,2304,256,5120,23040,61440,107520
%N Triangle read by rows: T(n,k) is the number of unordered pairs of vertices at distance k in the cube Q_n of dimension n (1 <= k <= n).
%C Sum of entries in row n = 2^(n-1)*(2^n-1) = A006516.
%C The entries in row n are the coefficients of the Wiener polynomial of the cube Q_n.
%C Sum_{k=1..n} k*T(n,k) = n*4^(n-1) = A002697(n) = the Wiener index of the cube Q_n.
%C Triangle T(n,k), 1 <= k <= n, read by rows given by [1,1,0,0,0,0,0,...]DELTA[1,1,0,0,0,0,0,...] where DELTA is the operator defined in A084938; subtriangle of triangle A055372. - _Philippe Deléham_, Oct 14 2008
%H Indranil Ghosh, <a href="/A143376/b143376.txt">Rows 1..125, flattened</a>
%H B. E. Sagan, Y-N. Yeh and P. Zhang, <a href="http://users.math.msu.edu/users/sagan/Papers/Old/wpg-pub.pdf">The Wiener Polynomial of a Graph</a>, Internat. J. of Quantum Chem., 60, 1996, 959-969.
%F T(n,k) = 2^(n-1)*binomial(n,k).
%F G.f.: G(q,z) = qz/((1-2z)(1-2z-2zq)).
%F T(n,k) = A055372(n,k). - _Philippe Deléham_, Oct 14 2008
%e T(2,1)=4, T(2,2)=2 because in Q_2 (a square) there are 4 distances equal to 1 and 2 distances equal to 2.
%e Triangle starts:
%e 1;
%e 4, 2;
%e 12, 12, 4;
%e 32, 48, 32, 8;
%e 80, 160, 160, 80, 16;
%p T:=proc(n,k) options operator, arrow: 2^(n-1)*binomial(n,k) end proc: for n to 10 do seq(T(n,k),k=1..n) end do; # yields sequence in triangular form
%t nn = 8; A[u_, z_] := (z + u z)/(1 - (z + u z));
%t Drop[Map[Select[#, # > 0 &] &, Map[Drop[#, 1] &,CoefficientList[Series[1/(1 - A[u, z]), {z, 0, nn}], {z, u}]]],1] // Grid (* _Geoffrey Critzer_, Mar 04 2017 *)
%t Flatten[Table[2^(n-1) Binomial[n, k], {n, 10},{k,n}]] (* _Indranil Ghosh_, Mar 06 2017 *)
%o (PARI) tabl(nn) = {for (n=1, nn, for(k=1, n, print1(2^(n-1) * binomial(n, k),", ");); print();); };
%o tabl(10); \\ _Indranil Ghosh_, Mar 06 2017
%o (Python)
%o import math
%o f=math.factorial
%o def C(n,r): return f(n) / f(r) / f(n-r)
%o i=1
%o for n in range(1,126):
%o ....for k in range(1,n+1):
%o ........print(str(i)+" "+str(2**(n-1)*C(n,k)))
%o ........i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y Cf. A002697, A006516.
%K nonn,tabl
%O 1,2
%A _Emeric Deutsch_, Sep 05 2008
%E Typo corrected by _Philippe Deléham_, Jan 05 2009
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